  Question

# Prove that the parallelogram circumscribing a circle is a rhombus. OR Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Solution

## Since ABCD is a parallelogram, AB = CD ...(1) BC = AD ...(2) It can be observed that DR = DS (Tangents on the circle from point D) CR = CQ (Tangents on the circle from point C) BP = BQ (Tangents on the circle from point B) AP = AS (Tangents on the circle from point A) Adding all these questions, we obtain DR + CR +BP + AP  =  DS + CQ + BQ + AS (DR + CR) + (BP + AP)  =  (DS + AS) + (CQ + BQ) CD + AB = AD + BC On putting the values of equations (1) and (2) in this equation, we obtain 2AB = 2BC AB = BC ...(3) Comparing equations (1),(2) and (3), we obtain AB = BC = CD = DAHence, ABCD is a rhombus. OR Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R, S. let us join the vertices of the quadrilateral ABCD to the center of the circle. Consider ΔOAP and ΔOAS, AP =AS (Trangents from the same point) OP =OS (Radii of the same circle) OA =OA (Common side) ΔOAP≅ΔOAS (SSS congruence criterion) THerefore, A↔A,P↔S,O↔O And thus, ∠POA=∠AOS ∠1=∠8 Similarly, ∠2=∠3∠4=∠5∠6=∠7 ∠1+∠2+∠3=∠4+∠5+∠6+∠7+∠8=360∘(∠1+∠8)+(∠2+∠3)+(∠4+∠5)+(∠6+∠7)=360∘2∠1+2∠2+2∠5+2∠6=360∘2(∠1+∠2)+2(∠5+∠6)=360∘(∠1+∠2)+(∠5+∠6)=180∘∠AOB+∠COD=180∘ Similarly, we can prove that ∠BOC+∠DOA=180∘ Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.  Suggest corrections   